First slide
Vertical projection from ground
Question

A juggler throws balls vertically up in the air. He throws one up whenever the previous one is at its highest point. If he throws n balls in each second, the height to which each ball rises above the point of throw will be (g is acceleration due to gravity)

Easy
Solution

The time interval is (1/n) s. In this time the ball rises/falls the required distance. Using s = ut + (1/2)at2, h = (1/2)g(1/n)2

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